# How to find a tangent to a circle from an external point using calculus?

So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .

• How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me. – amd Jan 12 at 9:32
• Why do we need calculus to begin with? Analytic geometry works perfectly well. – Oscar Lanzi Jan 12 at 11:11

Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...

• The OP asked for a solution that uses calculus. – amd Jan 12 at 9:23
• No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful. – Oscar Lanzi Jan 12 at 11:15

We maximize slope from a given fixed point $$(x_1,y_1)$$ to a parameterized circle variable point in its standard form:

$$(x-h) = a \cos t,\, (y-k)= a \sin t; \tag1$$ and solve for the parameter

$$\frac{y-y_1}{x-x_1}= \frac{k+ a \sin t -y_1}{h+ a \cos t -x_1}=\frac{-\cos t} {\sin t}\tag2$$

the latter has been obtained by chain rule differentiation

Simplifying get the condition to obtain $$t_1$$ of tangent points (two)

$$\sin t_1 (k-y_1)+ \cos t_1 (h-x_1)+a=0 \tag3$$

It can be also recognized as tangent-normal form of a straight line.

Let us call $$R$$ the ray and $$C = (x_0, y_0)$$ the center of the circle. The points of the circle are the $$M =(R\cos\theta, R\sin\theta)$$.
Let $$E = (a, b)$$ be the external point.
Then the two points $$M$$ corresponding to a tangeant passing through $$E$$ are such that vector $$\vec{CM}$$ is orthogonal to $$\vec{EM}$$.

It immediately gives: $$\cos\theta\,(x_0+ R\cos\theta - a)+ \sin\theta\,(y_0+R\sin\theta-b) = 0$$ And then $$\cos\theta\,(x_0-a) + \sin\theta\,(y_0-b) + R$$

One possibility to go further is to rewrite $$\vec{EC}$$ as $$(\rho\cos\phi, \rho\sin\phi )$$
And therefore $$\cos(\theta - \phi) = -\frac{R}{\rho}$$